# Aptitude - Surds and Indices - Discussion

Discussion Forum : Surds and Indices - General Questions (Q.No. 8)
8.
 1 + 1 + 1 = ? 1 + x(b - a) + x(c - a) 1 + x(a - b) + x(c - b) 1 + x(b - c) + x(a - c)
0
1
xa - b - c
None of these
Explanation:

Given Exp. =
1  +  1  +  1
 1 + xb + xc xa xa
 1 + xa + xc xb xb
 1 + xb + xa xc xc

 = xa + xb + xc (xa + xb + xc) (xa + xb + xc) (xa + xb + xc)

 = (xa + xb + xc) (xa + xb + xc)

= 1.

Discussion:
24 comments Page 2 of 3.

Franklin Emeka said:   8 years ago
I am not understanding this. Any of them can explain me?

Favour said:   8 years ago
I do not understand any of them.

Achitsa said:   9 years ago
First solve the denominator using LCM the answer you get. Find its reciprocal.

Akshita Tank said:   9 years ago
Can anybody explain the question in easier way?

KazirangaUniv said:   9 years ago
From the second step:

x^a(x^a+x^b+x^c)^2+x^b(x^a+x^b+x^c)^2+x^c(x^a+x^b+x^c)^2/(x^a+x^b+x^c)^3

Then take (x^a+X^b+x^c)^2 common from numerator and denominator and cut them both from top and bottom. Then we get step 3.

Shekhar kushwaha said:   9 years ago
What if x = 0?

Abhijeet singh said:   1 decade ago
In the second line just take the LCM. and make the adjustment accordingly.
Its very simple try it on paper itself.
You guys will get the answer.

Thank you.

How do you solve the second step? Really confusing, please do it the long way so we can understand.